Find the area of the smaller part of the circle x2+y2=a2 cut off by the line x=a/√2
Find the area of the smaller part of the circle x2+y2=a2 cut off by the line x=a/√2
Solution and Explanation
The area of the smaller part of the circle,x2+y2=a2,cut off by the line,x=a/√2,is the
area ABCDA.
It can be observed that the area ABCD is symmetrical about x-axis.
∴Area ABCD=2×Area ABC
Area of ABC=∫aa/√2ydx
=∫aa/√2√a2-x2dx
=[x/2√a2-x2+a2/2sin-1x/a]aa/√2
=[a2/2(π/2)-2a/2√2√a2-a2/2-a2/2sin(1/√2)]
=a2π/4-a/2√2.a/√2-a2/2(π/4)
=a2π/4-a2/4-a2π/8
=a2/4[π-1-π/2]
=a2/4[π/2-1]
⇒Area ABCD=2[a2/4(π/2-1)]=a2/2(π/2-1)
Therefore,the area of smaller part of the circle,x2+y2=a2,cut off by the line,x=a/√2,
is a2/2(π/2-1)units.
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